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Kryger [21]
2 years ago
14

Your heart pumps blood at a pressure of 100 mmHg and flow speed of 60 cm/s. At your brain, the blood enters capillaries with suc

h a large total cross-sectional area that the blood velocity is much smaller: about 0.7 mm/s. Ignore viscosity in the capillaries. How long could your neck be so that blood still reaches your brain in m
Physics
1 answer:
Savatey [412]2 years ago
7 0

Answer:

1.28 m

Explanation:

Generally, pressure of fluid is given by

P=\rho g h where g is acceleration due to gravity, h is the height and \rho is the density

Considering that the pressure for mercury is same as for blood only that the height and density of fluid are different then

\rho_b g h_b= \rho_m g h_m

Since g is constant, then

\rho_b h_b= \rho_m h_m

Making h_b the subject of the formula then

h_b=\frac {\rho_m h_m}{\rho_b}

Where subscripts m and b denote mercury and blood respectively

Assuming density of blood is 1060 Kg/m3, density of mercury as 13600 Kg/m3 and substituting height of mercury for 0.1 m then

h_b=\frac {13600*0.1}{1060}=1.283018868  m \approx 1.28 m

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An object is dropped from rest into a pit, and accelerates due to gravity at roughly 10 m/s2. It hits the ground in 5 seconds. A
vitfil [10]

Answer:

Second pit is 375 m deeper compared to first pit.

Explanation:

We have equation of motion s = ut + 0.5at²

First object hits the ground after 5 seconds,

          Initial velocity, u = 0 m/s

         Acceleration, a = 10 m/s²

         Time, t = 5 s

    Substituting,

                  s = ut + 0.5 at²

                 s = 0 x 5 + 0.5 x 10 x 5²

                    s = 125 m

           Depth of pit 1 = 125 m

Second object hits the ground after 10 seconds,

          Initial velocity, u = 0 m/s

         Acceleration, a = 10 m/s²

         Time, t = 10 s

    Substituting,

                  s = ut + 0.5 at²

                 s = 0 x 10 + 0.5 x 10 x 10²

                    s = 500 m

           Depth of pit 2 = 500 m

Difference in depths = 500 - 125 = 375 m

Second pit is 375 m deeper compared to first pit.

7 0
2 years ago
four children pull on the same stuffed toy at the same time , yet there is no net force on the toy.how is this possible?
Kay [80]
There was no net force on the stuffed toy, because the kids might have the same strength,  The same force is on both sides of it.  T<span>hey cancel each other out. They exert a force on the stuffed toy equal in strength but opposite in direction. The forces are balanced and the stuffed toy does not move.  </span>Its like a game of tug-o-war, but you and I have the same strength. the rope would be still and not moving. 
3 0
2 years ago
Read 2 more answers
What is the equation describing the motion of a mass on the end of a spring which is stretched 8.8 cm from equilibrium and then
Darya [45]

Answer:

y = -8.37 cm

Explanation:

As we know that the equation of SHM is given as

y = A cos(\omega t)

here we know that

\omega = \frac{2\pi}{T}

here we have

T = 0.66 s

now we have

\omega = \frac{2\pi}{0.66}

\omega = 3\pi

now we have

y = (8.8 cm) cos(3\pi t)

now at t = 2.3 s we have

y = (8.8 cm) cos(3\pi \times 2.3)

y = -8.37 cm

6 0
1 year ago
Read 2 more answers
Slick Willy is in traffic court (again) contesting a $50.00 ticket for running a red light. "You see, your Honor, as I was appro
Masteriza [31]

Answer:

61578948 m/s

Explanation:

λ_{actual} = λ_{observed} \frac{c+v_{o}}{c}

687 = 570 (\frac{3 * 10^{8} +v_{o} }{3 * 10^{8}} )

v_{o} = 61578948 m/s

So Slick Willy was travelling at a speed of 61578948 m/s to observe this.

8 0
2 years ago
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A jetboat is drifting with a speed of 5.0\,\dfrac{\text m}{\text s}5.0 s m ​ 5, point, 0, start fraction, start text, m, end tex
love history [14]

The question is incomplete. Here is the entire question.

A jetboat is drifting with a speed of 5.0m/s when the driver turns on the motor. The motor runs for 6.0s causing a constant leftward acceleration of magnitude 4.0m/s². What is the displacement of the boat over the 6.0 seconds time interval?

Answer: Δx = - 42m

Explanation: The jetboat is moving with an acceleration during the time interval, so it is a <u>linear</u> <u>motion</u> <u>with</u> <u>constant</u> <u>acceleration</u>.

For this "type" of motion, displacement (Δx) can be determined by:

\Delta x = v_{i}.t + \frac{a}{2}.t^{2}

v_{i} is the initial velocity

a is acceleration and can be positive or negative, according to the referential.

For Referential, let's assume rightward is positive.

Calculating displacement:

\Delta x = 5(6) - \frac{4}{2}.6^{2}

\Delta x = 30 - 2.36

\Delta x = - 42

Displacement of the boat for t=6.0s interval is \Delta x = - 42m, i.e., 42 m to the left.

8 0
2 years ago
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